Question

A particle initially moves in a horizontal circular path of radius π meters at a constant speed of 4 m/s.

With the initial speed of the particle as 4.0 m/s and changing at the rate of 5.0 m/s², the magnitude of the total angular displacement (in revolutions) of the particle after a time interval of 2 s is nearly?

Answer 1

Answer:

Angular displacement of the particle in 2 s is 5.73 rad

Explanation:

As we know that the angular speed of the particle is given as

$\omega = \frac{v}{R}$

so we have

$\omega = \frac{4}{\pi}$

so the centripetal acceleration of the particle is given as

$a_c = \omega^2 R$

Similarly the angular acceleration is given as

$\alpha = \frac{a_t}{R}$

$\alpha = \frac{5}{\pi}$

now in order to find the angular displacement we can use kinematics

$\theta = \omega t + \frac{1}{2}\alpha t^2$

$\theta = \frac{4}{\pi}(2) + \frac{1}{2}(\frac{5}{\pi})2^2$

$\theta = \frac{18}{\pi} rad = 5.73 rad$

#Learn

Topic : Circular motion

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